Further Results on an Equitable 1-2-3 Conjecture

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Further Results on an Equitable 1-2-3 Conjecture

Julien Bensmail, Foivos Fioravantes, Fionn Mc Inerney, Nicolas Nisse

To cite this version:

Julien Bensmail, Foivos Fioravantes, Fionn Mc Inerney, Nicolas Nisse.

Further Results on

an Equitable 1-2-3 Conjecture.

Discrete Applied Mathematics, Elsevier, 2021, 297, pp.1-20.

�10.1016/j.dam.2021.02.037 Get�. �hal-02533537v2�

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Further Results on an Equitable 1-2-3 Conjecture

∗

Julien Bensmail

1

_{, Foivos Fioravantes}

1

_{, Fionn Mc Inerney}

2

_{, and Nicolas Nisse}

1 1_{Université Côte d’Azur, Inria, CNRS, I3S, France}

2

Laboratoire d’Informatique et Systèmes, Aix-Marseille Université, CNRS, and Université de Toulon Faculté des Sciences de Luminy, Marseille, France

Abstract

In this work, we consider equitable proper labellings of graphs, which were recently intro-duced by Baudon, Pilśniak, Przybyło, Senhaji, Sopena, and Woźniak. Given a graph G, the goal is to assign labels to the edges so that 1) no two adjacent vertices are incident to the same sum of labels, and 2) every two labels are assigned about the same number of times. Partic-ularly, we aim at designing such equitable proper k-labellings of G with k being as small as possible. In connection with the so-called 1-2-3 Conjecture, it might be that labels 1, 2, 3 are, a few obvious exceptions apart, always sufficient to achieve this just as in the non-equitable version of the problem.

We provide results regarding some open questions about equitable proper labellings. Via a hardness result, we first prove that there exist infinitely many graphs for which more labels are required in the equitable version than in the non-equitable version. This remains true in the bipartite case. We finally show that, for every k ≥ 3, every k-regular bipartite graph admits an equitable proper k-labelling.

Keywords: Proper labellings, Equitable labellings, 1-2-3 Conjecture.

1 Introduction

This paper deals with a variant of the so-called 1-2-3 Conjecture, which is defined through the following notions. Consider a (simple undirected) graph G. Let us consider a k-labelling ` of G to be an assignment ` : E(G) → {1, . . . , k} of labels to the edges of G. For each vertex v of G, one can compute, as a colour c`(v) of v, the sum of the labels assigned by ` to the edges

incident to v. We say that ` is proper if c`is indeed a proper vertex-colouring of G, i.e., we have

c`(u) 6= c`(v) for every edge uv ∈ E(G). In this field, we say that G is nice if it has no connected

component isomorphic to K2. This is because K2 is indeed the only connected graph with no

proper labelling. In other words, the parameter χΣ(G), which is the minimum k such that G

admits a proper k-labelling, is defined for every nice graph G. Actually, Karoński, Łuczak, and Thomason conjectured in 2004 that χΣ(G) should never exceed 3 (see [6]):

Conjecture 1.1 (1-2-3 Conjecture). If G is a nice graph, then χΣ(G) ≤ 3.

A few facts on the 1-2-3 Conjecture are worth mentioning. In particular, the upper bound in the statement of the conjecture cannot be decreased to 2 in general, as shown for instance by complete

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graphs and odd-length cycles. More generally, it is known that, given a nice graph G, deciding whether χΣ(G) ≤ 2 holds is NP-complete (thus, unless P=NP, there is no “good” characterisation

of graphs admitting proper 2-labellings) [4]. This contrasts with the bipartite case, however, as it was more recently shown that bipartite graphs G with χΣ(G) = 3 form a restricted family of

graphs [12]. To date, the best result towards the 1-2-3 Conjecture was established by Kalkowski, Karoński, and Pfender in [5], who proved that χΣ(G) ≤ 5 holds for every nice graph G. We refer

the interested reader to [10] for a survey by Seamone on this topic.

Several side aspects of the 1-2-3 Conjecture are of interest, such as the importance of the three labels 1, 2, 3 for designing proper labellings. In particular, a total version of the 1-2-3 Conjecture introduced by Przybyło and Woźniak in [9], where also the vertices must be labelled, seems to indicate that, in general, for any nice graph we should be able to design 2-labellings that are close to proper. As a consequence, to the best of our knowledge there is no known (non-trivial) graph G with χΣ(G) = 3 for which we must use the three labels 1, 2, 3 with nearly equal proportion

in every proper 3-labelling. As an example, let us mention that for every nice complete graph Kn, for which χΣ(Kn) = 3, there is a proper 3-labelling assigning label 2 only once [2]. Also, for

every bipartite graph G with χΣ(G) = 3, there exist proper 3-labellings assigning label 3 at most

twice [3]. It might be that, in general, using three labels might be too powerful, in the sense that two labels are “almost enough”.

Our results in this work are related to a recent take on those questions by Baudon, Pilśniak, Przybyło, Senhaji, Sopena, and Woźniak. In [1], they investigated proper labellings in which all labels must be assigned about the same number of times. More precisely, given a labelling ` of a graph G, one can define, for every label α assigned by `, the value nb`(α) being the number of

edges of G assigned label α. We call ` equitable if, for every two distinct labels α, β assigned by `, the values nb`(α) and nb`(β) differ by at most 1. We denote by χΣ(G) the smallest k such that G

admits an equitable proper k-labelling. Again, this parameter χΣis defined for every nice graph.

The authors of [1] have investigated several aspects of equitable proper labellings, most of which are about the relationship between χΣ(G) and χΣ(G) for a given graph G. For a few families of

graphs G, they have notably established that χΣ(G) = χΣ(G) holds, except for a few exceptions.

In particular:

• For nice forests F , we always have χΣ(F ) = χΣ(F ) ≤ 2.

• For nice complete bipartite graphs Kn,m, we always have χΣ(Kn,m) = χΣ(Kn,m) ≤ 2, except

for the peculiar case of K3,3 which verifies 2 = χΣ(K3,3) < χΣ(K3,3) = 3.

• For nice complete graphs Kn, we always haveχΣ(Kn) = χΣ(Kn) = 3, except for the peculiar

case of K4which verifies 3 = χΣ(K4) < χΣ(K4) = 4.

At this point, the previous results lead to a number of interesting questions. Is K4 the only

graph G with χΣ(G) > 3? Are graphs G with χΣ(G) < χΣ(G) rare? Can the difference between

χΣ(G) and χΣ(G) be arbitrarily large? In general, is an equitable version of the 1-2-3 Conjecture

plausible?

Conjecture 1.2 (Equitable 1-2-3 Conjecture). If G is a connected graph different from K2 and

K4, then χΣ(G) ≤ 3.

A few more results partially answering some of these questions can be found in Senhaji’s thesis [11]. In particular:

• Senhaji proved that χΣ(G) = χΣ(G) ≤ 3 holds for a certain number of graphs G, including

nice paths, nice cycles, some Theta graphs, and some Cartesian products.

• Using computer programs, he came up with four cubic bipartite graphs G verifying 2 = χΣ(G) < χΣ(G) = 3.

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• For particular cubic bipartite graphs G, such as Hamiltonian ones, he proved that χΣ(G) ≤ 2.

In this work, we provide results towards some of the questions above. In particular, we in-vestigate the existence of graphs G with χΣ(G) < χΣ(G). In Section 2, we first prove that there

exist infinitely many such graphs. This is obtained through proving that the problem of deciding whether χΣ(G) = 2 holds for a given graph G with χΣ(G) = 2 is NP-complete. We then

investi-gate, in Section 3, the same question for bipartite graphs. We exhibit operations establishing that there exist infinitely many bipartite graphs G with χΣ(G) < χΣ(G). We also prove that for every

bipartite graph G with χΣ(G) = 3, we have χΣ(G) = 3. In Section 4, we finally provide a result

on equitable proper labellings of regular bipartite graphs, showing that χΣ(G) ≤ k holds for every

such k-regular graph (k ≥ 3). In particular, we have χΣ(G) ≤ 3 for every cubic bipartite graph G.

2 NP-completeness result

This section is devoted to the proof that the problem of deciding whether χΣ(G) = 2 holds for a

given graph G with χΣ(G) = 2 is NP-complete.

The reduction in the proof of our main result below will be mostly obtained by plugging several gadgets with specific properties together. More precisely, some of our gadgets will have specific pendent edges (i.e., with exactly one of their ends being of degree 1) being their inputs or outputs. Given two disjoint gadgets G and H where e is an output of G and f is an input of H, by plugging G and H (along e and f ) we mean identifying e and f together. More precisely, if e = xy and f = uv with y and v being the vertices of degree 1 of e and f respectively, then identifying e and f means identifying x and v, and y and u respectively.

Before proceeding with introducing the needed gadgets, let us first recall the following easy observation that will serve throughout this work.

Observation 2.1. Let G be a graph with a path (u, v, w, x) where d(v) = d(w) = 2. For any proper labelling ` of G, we have `(uv) 6= `(wx).

Proof. This is because we would have c`(v) = c`(w) otherwise.

2.1 Initiator gadget

The first gadget we need is the diamond D depicted in Figure 1. Here and further, for every gadget introduced through a figure, we deal with its vertices and edges following the notation from that figure. The input of D is the edge u1u2, while the output of D is the edge u9u10. The properties

of interest of D are the following: Theorem 2.2. D verifies the following:

• |E(D)| = 11.

• In any proper 2-labelling ` of D, we have `(u1u2) = `(u9u10).

• There exist both proper 2-labellings ` of D where `(u1u2) = 1, and proper 2-labellings ` of D

where `(u1u2) = 2.

• In any proper 2-labelling ` of D where `(u1u2) = 1:

– c`(u2) = 4;

– c`(u9) can be any value in {2, 3};

– nb`(1) = 7 and nb`(2) = 4.

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1 u1 4 u2 3 u3 5 u4 4 u5 2/3 u6 3/4 u7 4/3 u8 3/2 u9 1 u10 1 1 2 1 1 2 1 1/2 2 2/1 1 (a) `(u1u2) = 1. 2 u1 5 u2 4 u3 6 u4 5 u5 3/4 u6 2/3 u7 3/2 u8 4/3 u9 2 u10 2 1 2 2 1 2 2 1/2 1 2/1 2 (b) `(u1u2) = 2.

Figure 1: The diamond gadget D. The values in each vertex v are the possible colours of c`(v) by

a proper 2-labelling ` of D.

– c`(u2) = 5;

– c`(u9) can be any value in {3, 4};

– nb`(1) = 4 and nb`(2) = 7.

Proof. Let ` be a proper 2-labelling of D. Assume `(u1u2) = 1.

If `(u2u3) = `(u2u4) = 1, then c`(u2) = 3. Since c`(u3) 6= c`(u4), we have `(u3u5) = 1 and

`(u4u5) = 2, or vice versa. Since c`(u3) 6= c`(u2), we have `(u3u4) = 2. This gives c`(u3) = 4 and

c`(u4) = 5. Now we note that no matter what `(u5u6) is, we necessarily get c`(u5) ∈ {4, 5} =

{c`(u3), c`(u4)}, a contradiction. So we cannot have `(u2u3) = `(u2u4) = 1.

If `(u2u3) = `(u2u4) = 2, then c`(u2) = 5. Again, since c`(u3) 6= c`(u4), we have, say, `(u3u5) =

1 and `(u4u5) = 2. Since c`(u4) 6= 5, we have `(u3u4) = 2, which gives c`(u3) = c`(u2) = 5, a

contradiction. Thus, we cannot have `(u2u3) = `(u2u4) = 2.

If, say, `(u2u3) = 1 and `(u2u4) = 2, then c`(u2) = 4. Assume first that `(u3u4) = 2. In that

case, since c`(u3) 6= c`(u2), we have `(u3u5) = 2, and thus c`(u3) = 5. Since c`(u4) 6= c`(u3), we

have `(u4u5) = 2, and thus c`(u4) = 6. Now, we note that no matter what `(u5u6) is, we have

c`(u5) ∈ {5, 6} = {c`(u3), c`(u4)}, a contradiction. So, we have `(u3u4) = 1. Since c`(u3) and

c`(u4) are different from c`(u2), we have `(u3u5) = 1 and `(u4u5) = 2, which gives c`(u3) = 3

and c`(u4) = 5. Now, since c`(u5) 6∈ {3, 5} = {c`(u3), c`(u4)}, we have `(u5u6) = 1, which gives

c`(u5) = 4. By Observation 2.1, we then have `(u7u8) 6= `(u5u6) and `(u9u10) 6= `(u7u8), and thus

`(u7u8) = 2 and `(u9u10) = 1. By the same argument, we have `(u6u7) 6= `(u8u9), and both ways

are possible. Indeed, if on the one hand `(u6u7) = 1 and `(u8u9) = 2, then c`(u6) = 2, c`(u7) = 3,

c`(u8) = 4, and c`(u9) = 3. If on the other hand `(u6u7) = 2 and `(u8u9) = 1, then c`(u6) = 3,

c`(u7) = 4, c`(u8) = 3, and c`(u9) = 2. According to all these arguments, we have that nb`(1) = 7

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`(u8u9) = 1, the value of c`(u9) can be any one in {2, 3}. See Figure 1(a) for an illustration of the

resulting `.

These arguments can be mimicked the exact same way when `(u1u2) = 2. In particular, we

have `(u1u2) = `(u9u10) = 2, nb`(1) = 4 while nb`(2) = 7, and c`(u9) can be any value in {3, 4}.

See Figure 1(b) for an illustration.

The initiator gadget Ik of length k ≥ 2 has one input and one output, and is obtained from k

diamond gadgets as follows. For k = 2, the initiator gadget I2of length 2 is obtained by plugging

two copies D1and D2 of the diamond gadget D along the output of D1 and the input of D2. The

input of I2 is then the input of D1 and the output of I2is then the output of D2. For k > 2, the

initiator gadget Ik of length k is obtained by plugging a copy G of the initiator gadget Ik−1 of

length k − 1 and a new copy H of the diamond gadget D along the output of G and the input of H. The input of Ik is then the input of G and the output of Ik is then the output of H.

Theorem 2.3. Ik verifies the following, for every k ≥ 2:

• |E(Ik)| = 10k + 1.

• In any proper 2-labelling of Ik, the input and output are assigned the same label.

• There exist both proper 2-labellings of Ik where the input is assigned label 1, and proper

2-labellings of Ik where the input is assigned label 2.

• In any proper 2-labelling ` of Ik where the input is assigned label 1:

– c`(v) can be any value in {2, 3}, where v denotes the degree-2 vertex of the output of Ik;

– nb`(1) = 6k + 1 and nb`(2) = 4k.

• In any proper 2-labelling ` of Ik where the input is assigned label 2:

– c`(v) can be any value in {3, 4}, where v denotes the degree-2 vertex of the output of Ik;

– nb`(1) = 4k and nb`(2) = 6k + 1.

Proof. This follows mainly from the fact that Ik is made up of k copies of the diamond gadget D

plugged one after another, and that the diamond gadget D has all of the properties described in Theorem 2.2. In particular, a proper 2-labelling of Ik induces a proper 2-labelling of the k copies

of the diamond gadget D in it. Specifically, it can be checked that no conflict can arise around the inputs and outputs that were identified. Also, for a proper 2-labelling of Ik assigning label

α ∈ {1, 2} to the input, in each copy of the diamond gadget D label α must be assigned to 7 edges while the other label must be assigned to 4 edges. Due to how the copies of D were plugged, we deduce that 7k − (k − 1) = 6k + 1 edges of Ik must be assigned label α, while 4k edges must be

assigned the other label.

2.2 Corrector gadget

The corrector gadget C is the graph depicted in Figure 2. The input of C is the edge u1u2, while

C has no output. Its interesting properties are the following: Theorem 2.4. C verifies the following:

• |E(C)| = 9.

• There exist both proper 2-labellings ` of C where `(u1u2) = 1, and proper 2-labellings ` of C

where `(u1u2) = 2.

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1 u1 2 u2 3 u3 4 u4 4 u5 3 u6 5 u7 1 1 1 1 1 1 2 2 1 (a) `(u1u2) = 1, nb`(1) = 7, nb`(2) = 2. 1 u1 3 u2 6 u3 5 u4 5 u5 4 u6 6 u7 1 2 2 2 1 1 2 2 2 (b) `(u1u2) = 1, nb`(1) = 3, nb`(2) = 6. 2 u1 4 u2 6 u3 5 u4 5 u5 4 u6 6 u7 2 2 2 2 1 1 2 2 2 (c) `(u1u2) = 1, nb`(1) = 2, nb`(2) = 7.

Figure 2: The corrector gadget C. The values in each vertex v are the possible colours of c`(v) by

a proper 2-labelling ` of C.

– c`(u2) ∈ {2, 3};

– either nb`(1) = 7 and nb`(2) = 2, or nb`(1) = 3 and nb`(2) = 6.

• In any proper 2-labelling ` of C where `(u1u2) = 2:

– c`(u2) = 4;

– nb`(1) = 2 and nb`(2) = 7.

Proof. Let ` be a proper 2-labelling of C. Because u6 and u7 both have degree 3, we have that

c`(u6), c`(u7) ∈ {3, 4, 5, 6}. Furthermore, c`(u6) 6= c`(u7), and we cannot have {c`(u6), c`(u7)} =

{3, 6}. We consider all of the remaining possibilities for {c`(u6), c`(u7)} in what follows.

Assume c`(u6) = 3 and c`(u7) = 4. Then, all three edges incident to u6 must be labelled 1,

while we have, say, `(u7u4) = 1 while `(u7u5) = 2. Then, we note that, whatever `(u4u3) is,

we have c`(u4) ∈ {3, 4} = {c`(u6), c`(u7)}, a contradiction. The case where `(u7u4) = 2 while

`(u7u5) = 1 is symmetric with the colour of u5 coming into conflict with that of u6or u7 instead.

Thus, we cannot have {c`(u6), c`(u7)} = {3, 4}.

Assume c`(u6) = 3 and c`(u7) = 5. Again, all three edges incident to u6 must be labelled 1,

while we have `(u7u4) = `(u7u5) = 2. Now, since c`(u4) and c`(u5) are different from 5 = c`(u7),

we have `(u4u3) = `(u5u3) = 1. This gives c`(u4) = c`(u5) = 4. Now, since c`(u3) is different from

4 = c`(u4) = c`(u5), we have `(u3u2) = 1. Then c`(u3) = 3, and, since c`(u2) 6= c`(u3), we have

`(u2u1) = 1, which yields c`(u2) = 2. This is the labelling depicted in Figure 2(a).

Assume c`(u6) = 4 and c`(u7) = 5. First, assume `(u6u7) = 1. Then we have, say, `(u6u4) = 1

and `(u6u5) = `(u7u4) = `(u7u5) = 2 (the case where `(u6u4) = 2 and `(u6u5) = 1 is symmetric).

Note now that whatever `(u4u3) is, we have c`(u4) ∈ {4, 5} = {c`(u6), c`(u7)}, a contradiction.

Then, assume `(u6u7) = 2. Then we have `(u6u4) = `(u6u5) = 1 and, say, `(u7u5) = 1 and

`(u7u4) = 2 (the case where `(u7u5) = 2 and `(u7u4) = 1 is symmetric). Again, note that

whatever `(u4u3) is, we have c`(u4) ∈ {4, 5} = {c`(u6), c`(u7)}, a contradiction. Thus, we cannot

have {c`(u6), c`(u7)} = {4, 5}

while we have `(u6u4) = `(u6u5) = 1. Now, since c`(u4) and c`(u5) are different from 4 = c`(u6),

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2/3 a1 3/4 a2 2/3 b1 3/4 b2 8/9 u2 1/2 1/2 1 2 1 2 2/3 c1 3/4 c2 2/3 d1 3/4 d2 9/10 u3 1/2 1/2 1 2 1 2 3 u1 1 2 1/2

Figure 3: The triangle gadget T2. The values in each vertex v are the possible colours of c`(v) by

a proper 2-labelling ` of T2.

5 = c`(u4) = c`(u5), we have `(u3u2) = 2. Then, c`(u3) = 6, and note that u1u2 can correctly be

assigned either of the labels 1 and 2. In the first case, we get the labelling depicted in Figure 2(b), in which c`(u2) = 3. In the second case, we get the labelling depicted in Figure 2(c), in which

c`(u2) = 4.

while we have, say, `(u6u4) = 1 while `(u6u5) = 2 (the case where `(u6u4) = 2 while `(u6u5) = 1 is

symmetric). Then, we note that, whatever `(u5u3) is, we have c`(u5) ∈ {5, 6} = {c`(u6), c`(u7)},

a contradiction. Thus, we cannot have {c`(u6), c`(u7)} = {5, 6}.

2.3 Generator gadget

Let T2be the triangle gadget depicted in Figure 3, which has no input nor output. We call u1 the

root of T2. Note that T2 has 15 edges. It has the following labelling properties:

Theorem 2.5. In any proper 2-labelling ` of T2:

• {`(u1u2), `(u1u3)} = {1, 2};

• {c`(u2), c`(u3)} = {8, 9} or {c`(u2), c`(u3)} = {9, 10}; furthermore:

– if {c`(u2), c`(u3)} = {8, 9}, then nb`(1) can be any value in {6, . . . , 10};

– if {c`(u2), c`(u3)} = {9, 10}, then nb`(1) can be any value in {5, . . . , 9}.

Proof. Let ` be a proper 2-labelling of T2. Since c`(a1) 6= c`(a2), we have, say, `(a1u2) = 1 and

`(a2u2) = 2. Note that whatever `(a1a2) is, no conflict involving a1 (or a2) and u2 can arise,

due to the larger degree of u2. These arguments also apply around the bi’s, ci’s, and di’s. In

particular, the labels of the four edges joining u2 to the ai’s and bi’s bring 6 to the colour of u2,

and similarly the labels of the four edges joining u3 to the ci’s and di’s bring 6 to the colour of

u3. Now, since c`(u2) 6= c`(u3), we have, say, `(u1u2) = 1 and `(u1u3) = 2. Then, no conflict

involving u2 and u3 can arise, no matter whether u2u3 is labelled 1 or 2. In the first case, we get

(c`(u2), c`(u3)) = (8, 9), while we get (c`(u2), c`(u3)) = (9, 10) in the second case.

The parts of the statement dealing with nb`(1) hold, essentially, because each of the edges a1a2,

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1 u1 4 u2 5 u3 6 u4 10 u5 10 u6 11 u7 3/4 u8 2/3 u9 1 u10 3/4 u11 2/3 u12 1 u13 1 1 2 2 2 2 2 2 2 1/2 1 2 1/2 1 T2 T2 T2 T2 T2 (a) `(u1u2) = 1. 2 u1 5 u2 3 u3 4 u4 8 u5 8 u6 7 u7 2/3 u8 3/4 u9 2 u10 2/3 u11 3/4 u12 2 u13 2 1 2 1 1 1 1 1 1 1/2 2 1 1/2 2 T2 T2 T2 T2 T2 (b) `(u1u2) = 2.

Figure 4: The spreading gadget Gf_{. Every triangle with a “T}₂_{” inside indicates that a copy of the}

triangle gadget T2 is attached via its root. The values in each vertex v are the possible colours of

c`(v) by a proper 2-labelling ` of Gf.

The spreading gadget Gf is depicted in Figure 4. The input of Gf is its edge u1u2, while

its two outputs are its edges u9u10 and u12u13. Note that Gf contains five copies of the triangle

gadget T2 attached via their roots. The properties of interest of Gf are the following:

Theorem 2.6. Gf verifies the following: • |E(Gf_{)| = 89.}

• In any proper 2-labelling ` of Gf_{, we have `(u}₁_u₂_{) = `(u}₉_u₁₀_{) = `(u}₁₂_u₁₃_).

• There exist both proper 2-labellings ` of Gf _{where `(u}

1u2) = 1, and proper 2-labellings ` of

Gf _{where `(u}

1u2) = 2.

• In any proper 2-labelling ` of Gf _{where `(u}₁_u₂_{) = 1:}

– c`(u2) = 4;

– c`(u9) and c`(u12) can be any value in {2, 3}; furthermore:

∗ if c`(u9) = c`(u12) = 2, then nb`(1) can be any value in {35, . . . , 56};

∗ if {c`(u9), c`(u12)} = {2, 3}, then nb`(1) can be any value in {34, . . . , 55};

• In any proper 2-labelling ` of Gf _{where `(u}

1u2) = 2:

– c`(u2) = 5;

– c`(u9) and c`(u12) can be any value in {3, 4}; furthermore:

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∗ if {c`(u9), c`(u12)} = {3, 4}, then nb`(1) can be any value in {34, . . . , 55}.

Proof. Consider ` a proper 2-labelling of Gf_{. We first note that we have `(u}

3u5) = `(u4u6).

Indeed, suppose to the contrary that, e.g., `(u3u5) = 1 and `(u4u6) = 2 holds. Since there are

two copies of T2attached to u5, by Theorem 2.5, the colour of u5is 7 + `(u5u7) and it is adjacent

to a vertex with colour 9 (in T2). Similarly, because of the two copies of T2 attached to u6, the

colour of u6is 8 + `(u6u7) and it is adjacent to a vertex with colour 9 (in T2). Then, we must have

`(u5u7) = 1 and `(u6u7) = 2, so that c`(u5) = 8 and c`(u6) = 10. We also know that a neighbour

of u7from the graph T2attached to it has colour 9, and that this graph T2provides 3 to the colour

of u7 by Theorem 2.5. Then, u7 has colour 6 + `(u7u8) + `(u7u11), and the two edges u7u8 and

u7u11 must be labelled (with 1 or 2) in such a way that the colour of u7 does not meet any value

in {8, 9, 10}, which is impossible.

Now suppose `(u1u2) = 1, and consider the edges u2u3 and u2u4 (see Figure 4(a) for an

illustration). First, if `(u2u3) = `(u2u4), then note that ` cannot be proper according to the

arguments above since we would need to have `(u3u5) 6= `(u4u6) since c`(u3) 6= c`(u4). Thus,

`(u2u3) = 1 and `(u2u4) = 2 without loss of generality, and c`(u2) = 4. Note that, if `(u3u4) = 1,

then we necessarily get that c`(u3) or c`(u4) is equal to c`(u2) since we need `(u3u5) = `(u4u6).

Thus, `(u3u4) = 2. We then have `(u3u5) = 2 since c`(u3) 6= c`(u2), and also `(u4u6) = 2 since

c`(u4) 6= c`(u3) (and because `(u4u6) = `(u3u5) by the arguments above).

According to the arguments above, we have `(u5u7) = `(u6u7) = 2. By the same arguments

and since c`(u5) = c`(u6) = 10, we have `(u7u8) = `(u7u11) = 2. Then, `(u9u10) = `(u12u13) = 1

to avoid conflicts. Thus, assuming the input of Gf _{is labelled 1, then its two outputs are also}

labelled 1. A similar case analysis yields an analogous conclusion when `(u1u2) = 2, see Figure 4(b).

Let us conclude by pointing out that, in the proper labellings of Gf _{mentioned above, the only}

edges for which the assigned label can freely be either 1 or 2 are the edges u8u9, u11u12, 4 edges in

each of the four copies of T2attached to u5 and u6, and 5 edges in the copy of T2 attached to u7.

As pointed out earlier, all other edges must (up to symmetry) receive a particular label in {1, 2} as soon as that of u1u2 is fixed. It is then easy to check that the parts of the statement dealing

with nb`(1) are true.

The last gadget we need is the generator gadget Gmwith m ≥ 3 outputs and one input, which

is obtained from m − 1 spreading gadgets as follows. For m = 3, the generator gadget G3 with

3 outputs is obtained by plugging two copies H1 and H2 of the spreading gadget Gf along any

output of H1 and the input of H2. The input of G3 is then the input of H1 and the 3 outputs of

G3are the second (unplugged) output of H1and the two outputs of H2. For m > 3, the generator

gadget Gm with m outputs is obtained by plugging a copy G of the generator gadget Gm−1 with

m − 1 outputs and a new copy H of the spreading gadget Gf _{along one output of G and the input}

of H. The input of Gmis then the input of G and the m outputs of Gmare the remaining m − 2

(unplugged) outputs of G and the two outputs of H. Theorem 2.7. Gm verifies the following, for every m ≥ 3:

• |E(Gm)| = 88m − 87.

• In any proper 2-labelling of Gm, the input and m outputs are assigned the same label.

• There exist both proper 2-labellings of Gm where the input is assigned label 1, and proper

2-labellings of Gmwhere the input is assigned label 2.

• In any proper 2-labelling ` of Gmassigning label 1 to the input, nb`(1) ∈ {32m−31, . . . , 55m−

54}.

• In any proper 2-labelling ` of Gmassigning label 2 to the input, nb`(1) ∈ {33m−33, . . . , 56m−

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Proof. This follows essentially from Theorem 2.6, since Gmis made up of m − 1 copies of Gf. In

particular, any proper 2-labelling ` of Gm induces one of each of its m − 1 underlying Gf’s. As

pointed out in the statement of Theorem 2.6, vertices identified through the plugging operation cannot get in conflict.

The part of the statement dealing with nb`(1) is essentially because, in each copy of Gf in Gm,

there are 23 edges that can freely be set to 1 or 2 (4 edges in four attached copies of T2, 5 edges in

the last attached copy of T2, and 2 edges adjacent to the outputs). Assuming the input of Gm is

labelled 1 by `, according to Theorem 2.6 in each of the copies of Gfthe number of edges that can be assigned label 1 essentially ranges from 33 to 56. Thus, in Gm, the number of edges that can be

assigned label 1 ranges from 33(m − 1) − (m − 2) = 32m − 31 to 56(m − 1) − (m − 2) = 55m − 54. The computation is similar when the input of Gm is assigned label 2, the only difference is that

copies of Gf _{do not share edges labelled 1.}

2.4 Main result

We are now ready for the main result of this section.

Theorem 2.8. Given a graph G with χΣ(G) = 2, deciding if χΣ(G) = χΣ(G) is NP-complete.

Proof. The problem is clearly in NP, so we focus on proving it is NP-hard. We do it by reduction from the Monotone Cubic 1-in-3 SAT problem, which is NP-hard according to [8]. An instance of this problem is a 3CNF formula F in which every clause Cj= (x1∨x2∨x3) contains exactly three

distinct variables (not negated) and every variable xibelongs to exactly three distinct clauses. The

question is whether there is a 1-in-3 truth assignment to the variables of F , i.e., a truth assignment such that every clause has exactly one true variable. Given F , we construct, in polynomial time, a graph G such that F admits a 1-in-3 truth assignment φ if and only if G admits an equitable proper 2-labelling `.

High-level description. Before describing the explicit construction of G, let us first give some intuition about its desired structure, and how all the previous gadgets will be used.

The most important gadget for our construction is the generator gadget Gm, as the fact that

we can generate arbitrarily many pending edges with the same label by a proper 2-labelling is a very convenient feature. This, for instance, permits to make the colours of some vertices grow by a similar amount, or, as will be illustrated later, to forbid some values as vertex colours.

Gmhas several downsides, however. A first one is that we, a priori, do not know whether its

input and outputs will be labelled 1 or 2 by a proper 2-labelling. A second one is that Gm does

not comply well with equitability, in the sense that, generally speaking, it admits both proper 2-labellings highly favouring the number of assigned 1’s, and proper 2-labellings highly favouring the number of assigned 2’s.

To overcome these points, we will use the initiator gadget Ik and copies of the corrector gadget

C, in the following way. The initiator gadget Ik will be used to introduce a large imbalance in

favour of one of the two labels by any proper 2-labelling. By that, we mean an imbalance that is so big that even all the labelling freedom we have in Gmwill not allow to close the gap. To make

sure that the whole graph does admit equitable proper 2-labellings, however, we will add several copies of the corrector gadget C. The most important property of this gadget is that, in terms of equitability, its behaviour regarding label 1 and label 2 is far from symmetric. By that, we mean, as noted in Theorem 2.4, that the possibilities C grants highly depend on the label assigned to its input by a proper 2-labelling. If this label is 1, then we can both favour the number of assigned 1’s or favour the number of assigned 2’s. On the contrary, if this label is 2, then for sure the number of assigned 2’s is favoured.

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ej fj gj e0_j f_j0 g0j e00_j f_j00 gj00 cj

Figure 5: Structure around a clause vertex cj. Wiggly edges are outputs of Gµ.

By properly plugging an initiator gadget Ik (for a well chosen value of k) and corrector gadgets

C onto the generator gadget Gm, we can, in particular, make sure that the outputs of Gm are all

labelled 1 by an equitable proper labelling of the whole graph. This is because, by a proper 2-labelling assigning label 2 to the outputs, the initiator gadget Ikwould introduce a huge imbalance

in favour of the number of assigned 2’s, that is so huge that it cannot be caught up by the labelling freedom of Gmand the copies of the corrector gadget C.

Once we know that the input and all outputs of Gmmust be assigned 1 by an equitable proper

2-labelling, the forcing mechanisms in the whole graph then become much easier to track, and it then becomes easier to design an equivalence with a 1-in-3 truth assignment φ satisfying F . Precise details. The construction of G is as follows. Let us start from the cubic bipartite graph GF modelling the structure of the 3CNF formula F . That is, for every variable xi of F we add a

variable vertex vi to the graph, for every clause Cj of F we add a clause vertex cj to the graph,

and, whenever a variable xibelongs to a clause Cjin F , we add the formula edge vicjto the graph.

We also add a generator gadget Gµ with µ outputs to the graph, where µ = 10(42m + 30n)

(where, here and further, n is the number of variables in F and m is the number of clauses in F ) so that we have sufficiently many outputs on hand to perform what follows. We connect some of the outputs and make them adjacent to the clause and variable vertices as follows (see Figure 5 for an illustration for clause vertices):

• For every clause vertex cj:

– We first add three new vertices ej, fj, gj, joined via the edges ejfj, fjgj, and ejgj to form

a triangle. We now identify ej and the degree-1 vertex of each of 4 unused outputs of Gµ.

Similarly, we identify fj and the degree-1 vertex of each of 4 unused outputs of Gµ. We

next identify gj and the degree-1 vertex of each of 2 unused outputs of Gµ. We finally

add the edge gjcj to the graph.

– We then add three new vertices e0_j, f_j0, g0_j, forming a triangle. We then identify e0_j and 5 unused outputs of Gµ as above, fj0 and 5 unused outputs of Gµ, and gj0 and 3 unused

outputs of Gµ. We finally add the edge g0jcj to the graph.

– We finally add three new vertices e00_j, f_j00, g_j00, forming a triangle. We then identify e00_j and 7 unused outputs of Gµ as above, fj00 and 7 unused outputs of Gµ, and gj00 and 5 unused

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outputs of Gµ. We finally add the edge g00jcj to the graph.

• For every variable vertex vi:

– We first add three new vertices ri, si, ti joined to form a triangle (ri, si, ti, ri). We then

identify ri and 5 unused outputs of Gµ as above. Similarly, we then identify si and 5

unused outputs of Gµ. Then, we identify ti and 3 unused outputs of Gµ. Finally we add

the edge tivi to the graph.

– We then add three new vertices r_i0, s0_i, t0_i joined to form a triangle (r_i0, s0_i, t0_i, r_i0). We then identify r_i0 and 6 unused outputs of Gµ as above. Similarly, we then identify s0i and 6

unused outputs of Gµ. Then, we identify t0i and 4 unused outputs of Gµ. Finally we add

the edge t0_ivi to the graph.

– Finally, we identify vi with the degree-1 vertex of one unused output of Gµ.

Note that there are, at this point, a total of β = 3m + 12m + 8n = 15m + 8n edges in the graph that are not part of Gµ. More precisely, 3m of these edges are edges of GF, i.e., formula

edges, 12m of these edges are part or incident to the triangles added above and joined to the clause vertices, while 8n of these edges are part or incident to the triangles joined to the variable vertices. We refer to this graph, that is, the current one with β edges, and that contains none of the edges of Gµ, as G0F.

Note that since µ = 10(42m + 30n), then, at this point, only₁₀1µ outputs of Gµhave been used.

To each of the ₁₀9µ unused outputs of Gµ, we plug a new copy of the corrector gadget C. Now, we

add to the graph an initiator gadget Iα of length α that we plug to the input of Gµ, where α is

chosen to be the unique integer such that

2 ≤ ((6α + 1) + (32µ − 31)) − (4α + (56µ − 56) + β) ≤ 3.

The whole resulting graph is our G, whose input is the input of Iα. Clearly, G is obtained in

polynomial time from F .

Claim 2.9. Let ` be a proper 2-labelling of G. If the input is assigned label 2, then ` cannot be equitable.

Proof of the claim. Assume this is wrong, and consider ` an equitable proper 2-labelling of G assigning label 2 to the input. We investigate how many 1’s and 2’s must be assigned to several of the different gadgets that were plugged together to build G.

• Regarding the initiator gadget Iα of length α used in the construction of G, by Theorem 2.3

we get that ` must assign label 1 to exactly 4α edges and label 2 to exactly 6α + 1 edges. • By Theorem 2.7, in G the input of each of the 9

10µ copies of the corrector gadget C have their

input labelled 2 by `, since it coincides with an output of Gµ whose input is labelled 2 (by

Theorem 2.3, since the output of Iαand the input of Gµ coincide). By Theorem 2.4, in each

of these copies of C, there are 2 edges labelled 1 and 7 edges labelled 2.

Omitting all of the contributions of the corrector gadgets, we can state that there are, at this point, at least 2α + 1 more 2’s than 1’s. This imbalance must be fixed via the labelling of the other edges of Gµ (i.e., not the input of Gµ) and of G0F. By Theorem 2.7, at most 56µ − 56 edges of Gµ

can be assigned label 1 (which yield at least 32µ − 31 edges of Gµ labelled 2, due to the number

of edges of Gµ), while the number of edges of G0F is β. By our choice of α, it is then impossible

that the number of assigned 1’s by ` catches up with the number of assigned 2’s. This contradicts

the equitability of `.

Towards establishing the equivalence with a 1-in-3 truth assignment φ satisfying F , let us now see how a proper 2-labelling ` of G assigning label 1 to the input behaves. We start off by pointing out the following property of the triangles we have joined to the clause and variable vertices.

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Claim 2.10. Let γ ≥ 4. Let H be any graph with a triangle (u, v, w, u) and an edge xw (where x 6∈ {u, v}), and ` be a partial proper 2-labelling of H. Assume only the edges uv, vw, wu, and xw remain to be labelled, that the partial colour of u and v is γ − 2, and that the partial colour of w is γ − 4. Then, in every proper extension of ` to uv, vw, wu, and xw, we have `(uv) = 2, `(xw) = 1, and c`(w) = γ.

Proof of the claim. Since only uv, vw, wu, and xw remain to be labelled, and u and v currently have the same partial colour, so that c`(u) 6= c`(v) we have, say, `(uw) = 1 and `(vw) = 2. If

`(uv) = 1, then we get c`(u) = γ, c`(v) = γ + 1, while the partial colour of w is currently γ − 1. It

is then impossible to assign a correct label to xw, i.e., so that c`(w) 6∈ {γ, γ + 1} = {c`(u), c`(v)}.

So, we have `(uv) = 2, in which case c`(u) = γ + 1 and c`(v) = γ + 2. As above, the partial

colour of w is currently γ − 1, and, so that c`(w) 6∈ {γ + 1, γ + 2}, we must set `(xw) = 1. Then,

c`(w) = γ.

Claim 2.10, applied to the structure of G (and more precisely to that of G0_F), yields the following. Claim 2.11. For any proper 2-labelling ` of G assigning label 1 to the input:

• For each clause vertex cj, exactly one of its three incident formula edges is assigned label 1.

Hence, c`(cj) = 8.

• For each variable vertex vi, either all three of its incident formula edges are assigned label 1,

or they are all assigned label 2. Hence, c`(vi) ∈ {6, 9}.

• The number of edges in G0

F that are assigned label 1 by ` is 7m + 4n, while the number of

edges assigned label 2 is 8m + 4n.

Proof of the claim. Let ` be such a labelling of G. By Theorems 2.3 and 2.7, all outputs of Gm

must also be labelled 1 by `.

• Consider any clause vertex cj of G, and, in particular, the neighbouring triangle (ej, fj, gj).

Note that all the conditions are met to apply Claim 2.10. Similarly, this claim applies to the two triangles (e0_j, f_j0, g0_j) and (e00_j, f_j00, g00_j). From the claim, we get that `(gjcj) = `(g0jcj) =

`(g_j00cj) = 1, c`(gj) = 6, c`(g0j) = 7, and c`(g00j) = 9. Since cj is incident to only three other

edges, formula ones, one of them must be labelled 1 while the other two must be labelled 2 so that c`(cj) 6∈ {6, 7, 9}. Then, c`(cj) = 8.

• Consider any variable vertex vi of G. By the same arguments, we have `(tivi) = `(t0ivi) = 1,

c`(ti) = 7, and c`(t0i) = 8. Consequently, the three remaining (formula) edges incident

to vi must either all be labelled 1 by `, so that c`(vi) = 6, or all be labelled 2, so that

c`(vi) = 9 (recall that vi is also incident to an output of Gµ labelled 1). These are the only

two possibilities so that c`(vi) 6∈ {7, 8}.

The last part of the statement follows from Claim 2.10 and the arguments above. This concludes the proof. Note, in particular, that a consequence is that we have c`(cj) 6= c`(vi) for every clause

vertex cj and variable vertex vi.

Claim 2.11 gives us a direct equivalence between finding a proper 2-labelling of G where the input is labelled 1 and a 1-in-3 truth assignment to the variables of F . Indeed, consider a proper 2-labelling ` of G assigning label 1 to the input. We regard the fact that `(vicj) = 1 (respectively

`(vicj) = 2) as having, in F , variable vi bringing truth value true (respectively false) to clause Cj.

The condition in the first item of Claim 2.11 depicts the fact that, by a 1-in-3 truth assignment of F , a clause is considered satisfied only if it has exactly one true variable. The condition in the second item depicts the fact that, by a truth assignment, a variable brings the same truth value to all of the clauses that contain it. Thus, we can design a 1-in-3 truth assignment φ to the variables of F from `, and vice versa.

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Thus, F is 1-in-3 satisfiable if and only if G admits proper 2-labellings where the input is labelled 1. By Claim 2.9, all equitable proper 2-labellings of G (if any) must assign label 1 to the input. Thus, to prove that F is 1-in-3 satisfiable if and only if G admits equitable proper 2-labellings, it remains to show that G admits proper 2-labellings assigning label 1 to the input if and only if it admits equitable ones assigning label 1 to the input. Since every equitable proper labelling is a proper labelling, all that remains is to prove the following claim.

Claim 2.12. If G admits proper 2-labellings where the input is assigned label 1, then G admits equitable proper 2-labellings where the input is assigned label 1.

Proof of the claim. Let us consider a proper 2-labelling ` of G assigning label 1 to the input, obtained as follows. From Theorems 2.3 and 2.7, we know that all outputs of Gµ must also be

assigned label 1. We propagate ` in Iα, Gµ, and G0F while guaranteeing the following properties:

• In Iα, exactly 4α edges are assigned label 2 by `, while 6α + 1 edges are assigned label 1.

This is actually the only way to propagate ` in Iα, recall Theorem 2.3. Thus, here, there are

2α + 1 more assigned 1’s than assigned 2’s.

• In Gµ, the number of assigned 1’s is as small as possible, i.e., is 32µ − 31. In that case, the

number of assigned 2’s is 56µ − 56. Such a labelling can be achieved by Theorem 2.7. • In G0_F, the number of assigned 2’s is m more than the number of assigned 1’s. By Claim 2.11

and since G admits a proper 2-labelling where the input is assigned label 1, this property is attainable (while maintaining that the labelling is proper) and actually has to hold.

To summarise the above, at this point, the number of assigned 1’s is ((6α + 1) + (32µ − 31)) − (4α + (56µ − 56) + m) more than the number of assigned 2’s. Recall that α was chosen as the unique integer such that 2 ≤ ((6α + 1) + (32µ − 31)) − (4α + (56µ − 56) + β) ≤ 3. Thus, β − m + 2 ≤ ((6α + 1) + (32µ − 31)) − (4α + (56µ − 56) + m) ≤ β − m + 3, and hence, the number of assigned 1’s we have considered is either 14m + 8n + 2 or 14m + 8n + 3 more than the number of assigned 2’s (recall that β = 15m + 8n). It remains to consider the 9

10µ = 9(42m + 30n) copies

of the corrector gadget C in G. This means that the number of copies of C in G is much bigger than 14m + 8n + 3. By Theorem 2.4, we can propagate ` to some copies of C so that 6 edges are assigned label 2 and 3 edges are assigned label 1. This way, the number of assigned 2’s we have considered thus far catches up with the number of assigned 1’s.

For the remaining copies of C, we can assume that ` roughly alternates propagating following the two labelling possibilities described in Theorem 2.4 when the input is labelled 1, so that the number of assigned 2’s we have considered remains close yet slightly bigger than the number of assigned 1’s. If, eventually, ` is not equitable because the number of assigned 2’s is slightly bigger than the number of assigned 1’s, then we can freely switch from 2 to 1 the labels assigned to some edges of, e.g., triangles in the copies of T2 in some of the spreading gadgets Gf in Gµ. Recall

that all these edges are indeed currently assigned label 2 (since we have minimised the number of assigned 1’s in Gµ).

Something to take into consideration is that the labelling of Gµ we have considered above,

i.e., the one minimising the number of 1’s, does not comply with the two labelling schemes of the corrector gadget C. Indeed, when the spreading gadget Gf _{is labelled so that the input is}

labelled 1 and the number of 1’s is minimised, note that the vertices u9and u12must have colour 3,

which is not compatible with the labelling of C in Figure 2(b). In this case, it is necessary to make u9 (or u12) have colour 2 (so that they comply with the desired labelling of C) by just changing

to 1 the label of an incident edge labelled 2. This consequently makes the number of 2’s increase, which must be taken into consideration for deciding how to label the next copies of C.

Eventually, ` is equitable.

To finish off the proof, we prove that, regardless of whether F is 1-in-3 satisfiable, there always exist proper 2-labellings of G. In other words, we always have χΣ(G) = 2.

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Claim 2.13. There exist proper 2-labellings of G.

Proof of the claim. We show that G admits proper 2-labellings ` where the input is labelled 2. Recall that we do not care about equitability here. By Theorem 2.3, the output of Iα, which is the

input of Gµ, must be labelled 2 when its input is labelled 2. In turn, by Theorem 2.7, the outputs

of Gµ must be labelled 2 as well. Some of these outputs are the input of corrector gadgets. From

Theorems 2.3, 2.4, and 2.7, we get that there do exist partial proper 2-labellings ` of these gadgets in G such that no conflicts arise.

It remains to prove that such a partial labelling ` can properly be extended to the edges of G0

F. We demonstrate the arguments for one triangle (ej, fj, gj, ej) adjacent to a clause vertex cj,

but the arguments are identical regarding the other triangles and the variable vertices. Because ej and fj are incident to 4 edges of Gµ, which are all labelled 2 by the arguments above, ej and

fj already have partial colour 8. Let us assign label 2 to ejfj and fjgj, and label 1 to ejgj. This

gives c`(ej) = 11 6= c`(fj) = 12. At this point, gj has partial colour 7 (3 from the labelling of ejgj

and fjgj, and 4 from the two outputs of Gµ). Let us assign label 1 to gjcj, so that c`(gj) = 8.

Note that no two of ej, fj, and gj are in conflict, and also their colour is so big that no conflict

with adjacent vertices in Gµ can arise.

By repeating these arguments, for every clause vertex cj, its incident edges cjgj, cjgj0, and cjgj00

are labelled 1, while gj, gj0, and g00j have colour at least 8. In particular, cj has partial colour 3 at

this point. Similarly, for every variable vertex vi its incident edges vitiand vit0i are labelled 1, and

ti and t0ihave colour at least 10. Recall that an output of Gµ, which is labelled 2, is also attached

to vi. Then, the partial colour of viis 4 at this point. Let us finish the construction of the labelling

` by assigning label 1 to every formula edge. Since every clause vertex cj and variable vertex vi is

incident to exactly three such edges, we get that c`(cj) = 6 and c`(vi) = 7 for every clause vertex

cj and every variable vertex vi. Then, neither clause vertices nor variable vertices get in conflict

with any of their neighbours. In particular, no clause vertex gets in conflict with a variable vertex.

Then, ` is proper, as desired.

This concludes the proof.

Another interpretation of Theorem 2.8 is that, independently of whether Conjecture 1.2 is true or not, determining χΣ(G) is an NP-hard problem for a given graph G. It is also worth mentioning

that, in our reduction, the reduced graphs G we construct should always verify χΣ(G) ≤ 3. This

can be seen by noting that all gadgets and structures we have added to G themselves admit many equitable proper 3-labellings, some of which could possibly be combined to yield one of G.

3 Bipartite graphs G with χ

Σ

(G) < χ

Σ

(G)

In this section, we investigate the existence of bipartite graphs G with χΣ(G) < χΣ(G). In

Section 3.1, we first focus on bipartite graphs G with χΣ(G) = 3, as they stand as good candidates

of graphs that could have χΣ(G) > 3. We prove that, actually, χΣ(G) = 3 holds for all these

graphs. In Section 3.2, we then study the existence of bipartite graphs G with χΣ(G) = 2 and

χΣ(G) = 3. We provide operations for building infinitely many such graphs.

3.1 Bipartite graphs G with χ

Σ

(G) = 3

A precise characterisation of bipartite graphs G with χΣ(G) = 3 was provided by Thomassen, Wu,

and Zhang in [12], where they proved that these graphs, called odd multi-cacti, can be obtained as follows. The definition is recursive.

• Any cycle C4k+2 with length 4k + 2 (k ≥ 1) with its edges being properly coloured with red

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• Given an odd multi-cactus G with edges coloured red and green, another odd multi-cactus is obtained when considering any green edge uv of G, and joining uv by a new path P of length 4k + 1 (k ≥ 1), where the edges of P are properly coloured with red and green in such a way that the two end-edges (i.e., incident to u and v) are coloured red.

In the next result, we prove that for every odd multi-cactus G, we have χΣ(G) = χΣ(G) = 3.

Theorem 3.1. If G is an odd multi-cactus, then χΣ(G) = 3.

Proof. The proof is by induction on the number of vertices of G. The base case corresponds to G being C6, the cycle of length 6, which is the smallest odd multi-cactus. We first prove a more

general case, namely that the claim is true whenever G is a cycle with length at least 6 congruent to 2 modulo 4.

Let G be a cycle with length at least 6 congruent to 2 modulo 4. In this case, an equitable proper 3-labelling can be obtained as follows. Traverse the successive edges of G starting from an arbitrary one, and assign labels 1, 2, 3, 1, 2, 3, . . . going along until all edges are labelled. Note that, doing so, at any moment of the procedure the resulting partial labelling is equitable.

• If the length of G is congruent to 0 or 1 modulo 3, then, by Observation 2.1, the resulting labelling is proper. This is because no two edges at distance 2 receive the same label, which is the only colour conflict that can occur in a path.

• If the length of G is congruent to 2 modulo 3 (the smallest such graph is C14), then we get

a conflict because of the last two edges that were labelled l1 = 1 and l2 = 2 respectively,

which are each at distance 2 from an edge with the same label. In this situation, we change l1

into 3 and l2 into 1. Note that no conflict remains now. Furthermore, the labelling remains

equitable (the number of assigned 2’s is one less than the numbers of assigned 1’s and 3’s, which are equal).

For the general case: suppose that all odd multi-cacti with order at most some x − 1 admit an equitable proper 3-labelling, and let us consider odd multi-cacti with order x. If x 6≡ 2 mod 4, then, by construction, there exist no such graphs on x vertices, and the claim is true. Thus, we assume that x ≡ 2 mod 4.

Let G be an odd multi-cactus with x vertices. Since G can be assumed to be different from a cycle, it was obtained from a cycle of length 2 modulo 4 by repeated path attachments onto green edges. Due to the structure of G, it can be noted that there has to exist a green edge uv where:

1. there exist p ≥ 1 disjoint paths P1, . . . , Pp joining u and v, all of whose inner vertices have

degree 2, and

2. the graph G0 obtained by removing the inner vertices of the paths P1, . . . , Pp from G is an

odd multi-cactus where u and v have degree 2.

First off, it can be assumed that all of the Pi’s have length 5. This is a consequence of the

following more general result:

Claim 3.2. Let P9 = (v1, . . . , v9) be the path of length 8, and assume we are given a partial

proper 3-labelling `0 of P9where only the four edges v1v2, v2v3, v7v8, and v8v9 are labelled, so that

`0(v1v2) 6= `0(v7v8) and `0(v2v3) 6= `0(v8v9). Then, for any permutation {α, β, γ} of {1, 2, 3}, it is

possible to extend `0 to a proper 3-labelling ` of P9 where two of v3v4, v4v5, v5v6, and v6v7 are

labelled α, one of these edges is labelled β, and one of these edges is labelled γ.

Proof of the claim. So that a labelling of P9is proper, we must only ensure that every two edges at

distance 2 receive distinct labels, recall Observation 2.1. In particular, this implies that labelling v3v4 and v5v6 can be done independently from labelling v4v5 and v6v7. Then, we will be done if

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we can prove that labels α and β can correctly be assigned to v3v4and v5v6, while labels α and γ

can correctly be assigned to v4v5and v6v7.

Without loss of generality, let us assume we want to assign labels α and β to v3v4 and v5v6.

Note that, at this point, `(v3v4) must only differ from `(v1v2). Let us assume that we can assign

`(v3v4) = α without there being a conflict with `(v1v2), i.e., `(v1v2) 6= α. If no conflict arises upon

setting `(v5v6) = β, then we are done. Otherwise, it means `(v7v8) = β. In that situation, let us

instead set `(v5v6) = α and `(v3v4) = β. If this raises a conflict, this must be because `(v1v2) = β.

But then, we deduce that `(v1v2) = `(v7v8) = β, a contradiction.

Indeed, assume, without loss of generality, that P1 has length 4k + 1 for some k ≥ 2. Let us

denote by (u, v1, . . . , v4k, v) the successive vertices of P1from u to v. Let G0 be the graph obtained

from G by removing the edges v2v3, v3v4, v4v5, v5v6, v6v7 and joining the vertices v2 and v7 by an

edge e. Note that G0 is an odd multi-cactus since we have essentially contracted a path of length 4k +1 (with k ≥ 2) into a path of length 4(k −1)+1. Then, by the induction hypothesis, there is an equitable proper 3-labelling `0 of G0. By definition, note that `0(uv1) 6= `0(e), `0(e) 6= `0(v8v9) (or

`0(e) 6= `0(v8v) if k = 2), and `0(v1v2) 6= `0(v7v8). To extend `0 to an equitable proper 3-labelling

` of G, for every edge that is both in G and G0 we first infer the label by `0 to `. We then set `(v2v3) to `0(e). Note that no conflict arises in G by this partial labelling. Furthermore, since

`0 is equitable, so is `. Now, consider {α, β, γ} a permutation of {1, 2, 3} such that every two of nb`0(α) + 2, nb_`0(β) + 1, and nb_`0(γ) + 1 differ by at most 1. Such an {α, β, γ} exists since `0 is

equitable. By Claim 3.2, the current ` can be extended to the edges v3v4, v4v5, v5v6, v6v7 so that

a proper 3-labelling of G results, and that this can be done by assigning label α twice, and each of β and γ once. By our choice of α, β, γ, such a resulting labelling is also equitable.

Then, we can assume all Pi’s have length exactly 5. For every i ∈ {1, . . . , p}, let us set

Pi = (u, vi1, . . . , vi4, v). Let us also denote by u0 the second neighbour (different from v) of u in

G0, and by v0 the second neighbour (different from u) of v in G0. Our goal is to extend `0 to the edges of the Pi’s so that no conflict arises, and the resulting 3-labelling ` of G is proper. To

begin, consider {α, β, γ} a permutation of {1, 2, 3}. The choice of α, β, and γ can be done in such a way that the ensuing labelling ` is equitable. Precisely, if nb`0(1) = nb_`0(2) = nb_`0(3) or

nb`0(1) + 1 = nb_`0(2) = nb_`0(3) or nb_`0(1) = nb_`0(2) = nb_`0(3) − 1, then (α, β, γ) = (1, 2, 3), else if

nb`0(1) − 1 = nb_`0(2) = nb_`0(3) or nb_`0(1) = nb_`0(2) + 1 = nb_`0(3), then (α, β, γ) = (2, 3, 1), and if

nb`0(1) = nb_`0(2) = nb_`0(3) + 1 or nb_`0(1) = nb_`0(2) − 1 = nb_`0(3), then (α, β, γ) = (3, 1, 2). For all

1 ≤ i ≤ p and for any a, b ∈ Pi such that ab ∈ E(G), it is easy to verify that c`(a) 6= c`(b), for any

of the labellings ` proposed below.

In what follows, let x be any vertex in X = N (u)

p

T

i=1

Pi, let y be any vertex in Y = N (v) p

T

i=1

Pi,

and let w be any vertex in W = (N (u) ∪ N (v))

p

T

i=1

Pi. Also, for all 1 ≤ i ≤ p, let `(Pi) =

(`(uv1i), `(v1iv2i), `(vi2vi3), `(v3iv4i), `(vi4v)).

Case p = 2:

All the possible subcases are illustrated in Table 1. Note that, in all of these subcases, the labelling ` has the property that P

x∈X

`(ux) = P

y∈Y

`(vy), and so, c`(u) 6= c`(v) since c`0(u) 6= c_`0(v)

by the inductive hypothesis. Furthermore, note that the maximum colour of any vertex w ∈ W is 6, no matter the labelling. Lastly, it may seem that the subcase c`0(u) + α + β 6= c_`0(u0),

c`0(v) + α + β = c_`0(v0) has not been treated, but it is actually symmetric to the subcase where

c`0(u) + α + β = c_`0(u0), c_`0(v) + α + β 6= c_`0(v0), which has been treated through subcases 2-4 in

Table 1. Case p ≥ 3:

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Subcase Labelling of No conflicts between u (v resp.) conditions paths P1 and P2 and any of its neighbours in P1 or P2

c`0(u) + α + β 6= c_`0(u0), c`0(v) + α + β 6= c_`0(v0) `(P1) = (α, β, γ, α, β) `(P2) = (β, α, γ, γ, α) If α = 1: c`(w) ≤ 4 & c`(u), c`(v) ≥ 5. If α = 2: c`(u), c`(v) ≥ 7. If α = 3: c`(w) ≤ 5 & c`(u), c`(v) ≥ 6. c`0(u) + α + β = c_`0(u0), c`0(v) + 2α 6= c_`0(v0), c`0(v) + 2α 6= α + γ `(P1) = (α, β, β, γ, α) `(P2) = (α, β, γ, γ, α) If α = 1: c`(x) = 3 & c`(u) ≥ 4. If α = 2: c`(x) = 5 & c`(u) ≥ 6. If α = 3: c`(u) ≥ 8. c`0(u) + α + β = c_`0(u0), c`0(v) + 2α 6= c_`0(v0), c`0(v) + 2α = α + γ `(P1) = (α, γ, β, β, α) `(P2) = (α, γ, γ, β, α)

Since c`0(u) 6= c_`0(v), then

c`0(u) + 2α 6= α + γ. c`0(u) + α + β = c_`0(u0), c`0(v) + 2α = c_`0(v0) `(P1) = (γ, α, β, γ, α) `(P2) = (α, γ, β, β, γ) If α = 1: c`(w) ≤ 5 & c`(u), c`(v) ≥ 6. If α = 2: c`(w) ≤ 4 & c`(u), c`(v) ≥ 5. If α = 3: c`(u), c`(v) ≥ 7. Table 1: The four subcases for p = 2.

Give the same labellings as in the case p = 2 for P1 and P2. For the remainder of the paths,

simply label them so that the labelling ` is equitable (and proper) and so that, for all 3 ≤ j ≤ p, we have `(uv₁j) = `(vvj₄). Note that, in this case, for all x ∈ X, it is not possible for c`(u) = c`(x).

Indeed, if c`(u) ≥ 7, then we are done since, for all w ∈ W , we have that c`(w) ≤ 6. Otherwise,

if c`(u) = 6, then, for all x ∈ X, we have that `(ux) ≤ 2, and so, c`(x) ≤ 5. Lastly, if c`(u) = 5

(note that this is the last case since c`(u) ≥ 5), then, for all x ∈ X, we have that `(ux) = 1, and

so, c`(x) ≤ 4. The same can be said for all y ∈ Y and v.

Case p = 1:

All the possible subcases are illustrated in Table 2. Note that, in the first six of these subcases, the labelling ` has the property that `(uv1) = `(vv4), and so, c`(u) 6= c`(v) since c`0(u) 6= c_`0(v)

by the inductive hypothesis. Furthermore, note that the logical subcase that would follow the last subcase in Table 2 would be that c`0(u) + α = c_`0(u0), c_`0(v) + β = c_`0(v0), c_`0(u) + β 6= c_`0(v) + α,

c`0(v)+α = 2α, and c_`0(u)+β = β +α. However, this subcase cannot exist since, if c_`0(v)+α = 2α,

then c`0(v) = α, and if c_`0(u) + β = β + α, then c_`0(u) = α, and hence, we have that c_`0(u) = c_`0(v),

a contradiction. Lastly, it may seem that the subcase c`0(u) + α 6= c_`0(u0), c_`0(v) + α = c_`0(v0) has

not been treated, but again, it is actually symmetric to the subcase where c`0(u) + α = c_`0(u0),

c`0(v) + α 6= c_`0(v0), which has been treated through subcases 4-12 in Table 2. This concludes the

proof as all of the possible cases have now been covered.

3.2 Bipartite graphs G with χ

Σ

(G) = 2

We start by introducing two operations, Operations 1 and 2, which, when applied to graphs G with χΣ(G) ≥ 3, provide more graphs G0 with χΣ(G0) ≥ 3.

Theorem 3.3 (Operation 1). Let G be a multigraph with χΣ(G) ≥ 3. If G has an edge uv with

multiplicity at least 2, then the graph G0 obtained from G by subdividing one of these edges uv four times verifies χΣ(G0) ≥ 3.

Proof. Let G0be obtained from G by replacing one edge uv with a path (u, w, x, y, z, v) of length 5. Assume there exists `0, an equitable proper 2-labelling of G0. Assume that `0(uw) = α for some

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Subcase Labelling No conflicts between u (v resp.) conditions of path P1 and any of its neighbours in P1

c_`0(u) + α 6= c_`0(u0), c_`0(v) + α 6= c_`0(v0), c_`0(v) + α 6= β + α, c_`0(u) + α 6= γ + α

`(P1) = (α, γ, β, β, α) By the conditions of the subcase.

c_`0(u) + α 6= c_`0(u0), c`0(v) + α 6= c`0(v0),

c_`0(v) + α 6= β + α, c_`0(u) + α = γ + α

`(P1) = (α, β, β, γ, α) Since c`0(u) 6= c`0(v), then c`0(v) + α 6= γ + α.

c_`0(u) + α 6= c_`0(u0), c_`0(v) + α 6= c_`0(v0_),

c_`0(v) + α = β + α

`(P1) = (α, β, β, γ, α)

Since c`0(u) 6= c`0(v), then

c_`0(u) + α 6= β + α. c_`0(u) + α = c_`0(u0),

c`0(v) + β 6= c`0(v0),

c_`0(v) + β 6= β + γ, c`0(u) + β 6= β + α

`(P1) = (β, α, α, γ, β) By the conditions of the subcase.

c_`0(u) + α = c_`0(u0), c_`0(v) + β 6= c_`0(v0_),

c_`0(v) + β 6= β + γ, c_`0(u) + β = β + α

`(P1) = (β, γ, α, α, β)

Since c_`0(u) 6= c_`0(v), then c_`0(v) + β 6= β + α. c`0(u) + α = c`0(u0),

c_`0(v) + β 6= c_`0(v0), c`0(v) + β = β + γ

`(P1) = (β, γ, α, α, β)

Since c_`0(u) 6= c_`0(v), then c_`0(u) + β 6= β + γ. c_`0(u) + α = c_`0(u0), c_`0(v) + β = c_`0(v0), c_`0(u) + β = c_`0(v) + α, c_`0(v) + γ 6= γ + α, c_`0(u) + β 6= 2β `(P1) = (β, β, α, α, γ)

By the conditions of the subcase. Note also that c`(u) 6= c`(v).

c_`0(u) + α = c_`0(u0), c`0(v) + β = c`0(v0), c_`0(u) + β = c_`0(v) + α, c_`0(v) + γ 6= γ + α, c_`0(u) + β = 2β `(P1) = (γ, α, β, β, α)

Since c_`0(u) = β, then c`0(v) = 2β − α, c`0(u0) = β + α,

c_`0(v0) = 3β − α, c`(u) = β + γ, and

c`(v) = 2β. It can then be verified

that there are no conflicts. c_`0(u) + α = c_`0(u0_),

c_`0(v) + β = c_`0(v0), c`0(u) + β = c`0(v) + α,

c_`0(v) + γ = γ + α

`(P1) = (β, α, α, β, γ)

By the last two conditions of the subcase, c_`0(u) + β = 2α.

Thus, c`(u) 6= β + α.

Note also that c`(u) 6= c`(v).

c_`0(u) + α = c_`0(u0), c_`0(v) + β = c_`0(v0), c_`0(u) + β 6= c_`0(v) + α, c_`0(v) + α 6= 2α, c_`0(u) + β 6= 2β `(P1) = (β, β, γ, α, α)

c_`0(u) + α = c_`0(u0_), c_`0(v) + β = c_`0(v0), c`0(u) + β 6= c`0(v) + α, c_`0(v) + α 6= 2α, c`0(u) + β = 2β `(P1) = (β, α, γ, β, α)

Note that β ≥ 2 since c`0(u) = β.

If β = 2: c`(u) = 4, c`(v) = 5, c`(v0) = 6 & c`(u0) = c`(x) = c`(y) = 3. If β = 3: c`(u) = 6, c`(v) = 4, c`(v0) = 7 & c`(u0) = c`(x) = c`(y) = 5. c_`0(u) + α = c_`0(u0), c_`0(v) + β = c_`0(v0), c_`0(u) + β 6= c_`0(v) + α, c_`0(v) + α = 2α, c_`0(u) + β 6= β + α `(P1) = (β, α, γ, β, α)

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v1 v2 v3 v4 v5 v6 1 1 2 2 2 2 1 2 2

(a) With parallel edges.

v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 2 2 2 2 2 2 1 1 1 1 2 2 1

(b) With adjacent degree-2 vertices.

Figure 6: Proper 2-labellings of two bipartite graphs G with 2 = χΣ(G) < χΣ(G) = 3.

α ∈ {1, 2}, and that `0(wx) = β for some β ∈ {1, 2}. Set {α} = {1, 2} \ {α} and {β} = {1, 2} \ {β}. Then, by Observation 2.1, we have `0(xy) = α, `0(yz) = β, and `0(zv) = α. By the properness of `0, since u and v are adjacent in G0, we have that c`0(u) 6= c_`0(v). This implies that the 2-labelling ` of

G obtained from `0by assigning label α to the edge uv that was subdivided for constructing G0, and setting `(e) = `0(e) for every e ∈ E(G) ∩ E(G0), is proper. Furthermore, we have {`0(wx), `0(yz)} = {`0_{(xy), `}0_{(zv)} = {1, 2}. Hence, nb}

`(1) = nb`0(1)−2 and nb_`(2) = nb_`0(2)−2. So, ` is an equitable

proper 2-labelling of G, a contradiction. Thus, χΣ(G0) ≥ 3.

Theorem 3.4 (Operation 2). Let G be a graph with χΣ(G) ≥ 3. If G has an edge uv with d(u) =

d(v) = 2, then the graph G0 obtained from G by subdividing uv four times verifies χΣ(G0) ≥ 3.

Proof. Let us denote by (u, w, x, y, z, v) the path of length 5 of G0 that results from the subdivision of uv. Let us denote the other neighbour of u by u0, and the other neighbour of v by v0. Assume there exists `0, an equitable proper 2-labelling of G0. Assume that `0(u0u) = α for some α ∈ {1, 2},

and that `0(uw) = β for some β ∈ {1, 2}. Set {α} = {1, 2} \ {α} and {β} = {1, 2} \ {β}.

According to Observation 2.1, we have `0(wx) = α, `0(xy) = β, `0(yz) = α, `0(zv) = β, and `0(vv0) = α. It follows that the 2-labelling ` of G obtained from `0 by setting `(uv) = β and `(e) = `0(e) for every e ∈ E(G) ∩ E(G0) is proper. In particular, note that the above implies that c`(u) = c`0(u) 6= c_`0(v) = c_`(v), although u and v are not adjacent in G0. Furthermore,

we have {`0(wx), `0(yz)} = {`0(xy), `0(zv)} = {1, 2}. This implies that nb`(1) = nb`0(1) − 2

and nb`(2) = nb`0(2) − 2. So, ` is an equitable proper 2-labelling of G, a contradiction. Thus,

χΣ(G0) ≥ 3.

We note in particular that Operations 1 and 2 mentioned in Theorems 3.3 and 3.4, when performed on bipartite graphs, yield graphs that are also bipartite. From this observation, we come up with two infinite families of bipartite graphs G verifying 2 = χΣ(G) < χΣ(G) = 3.

The first such family is obtained by repeatedly applying Operations 1 and 2 from the cubic bipartite multigraph depicted in Figure 6(a). This graph indeed has the following properties: Theorem 3.5. Let G be the cubic bipartite multigraph depicted in Figure 6(a). Then, χΣ(G) = 2

and χΣ(G) = 3.

Proof. Since G is cubic, we have χΣ(G) > 1 and thus, χΣ(G) > 1. Actually, we even have χΣ(G) =

2 since G does not match the definition of an odd multi-cactus (a proper 2-labelling is also included in Figure 6(a)). Towards a contradiction, assume G admits an equitable proper 2-labelling `. In

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order to have c`(v1) 6= c`(v2), we have `(v1v3) 6= `(v2v4). Similarly, since c`(v5) 6= c`(v6), we have

`(v5v3) 6= `(v6v4). Now, since c`(v3) 6= c`(v4), we have `(v1v3) = `(v3v5) and `(v2v4) = `(v4v6),

since otherwise, c`(v3) = c`(v4) = 1 + 2 + `(v3v4) by the previous argument. Thus, without loss of

generality, we may assume that `(v1v3) = `(v3v5) = 1 and `(v2v4) = `(v4v6) = 2.

Assume now that `(v3v4) = 1. This gives c`(v3) = 3 and c`(v4) = 5. Now, note that the two

edges joining v1 and v2, and similarly the two edges joining v5 and v6, cannot both be assigned

label 1 (as, otherwise, v1or v5would be in conflict with v3). Similarly, to avoid a conflict with v4,

the two edges joining v1and v2, and similarly the two edges joining v5and v6, cannot be assigned

labels 1 and 2 respectively. Thus, these four edges must be assigned label 2, which means that nb`(1) = 3 and nb`(2) = 6. This is a contradiction to the equitability of `. Similar arguments can

be used to show that we cannot have `(v3v4) = 2 either. Thus, χΣ(G) > 2, and one can easily

come up with equitable proper 3-labellings of G.

A second infinite family of bipartite graphs G with 2 = χΣ(G) < χΣ(G) = 3 is obtained by

repeatedly applying Operation 2, described in Theorem 3.4, to the graph depicted in Figure 6(b) (note that this is K3,3with one edge subdivided 4 times), which indeed has the following properties:

Theorem 3.6. Let G be the subcubic bipartite graph depicted in Figure 6(b). Then, χΣ(G) = 2

and χΣ(G) = 3.

Proof. Again, G is not locally irregular (i.e., it has adjacent vertices with the same degree) and does not match the structure of an odd multi-cactus, so χΣ(G) = 2 (a proper 2-labelling is also

included in Figure 6(b)). We now prove that χΣ(G) = 3. Let us suppose that there exists an

equitable proper 2-labelling ` of G. By Observation 2.1, we must have `(v1v10) 6= `(v9v8) and

`(v7v6) 6= `(v9v8), and thus, `(v1v10) = `(v7v6). Moreover, either `(v10v9) = `(v1v10) = `(v7v6)

and `(v9v8) = `(v8v7) or `(v1v10) = `(v8v7) = `(v7v6) and `(v10v9) = `(v9v8). This implies that

each of the labels 1 and 2 appears exactly twice in the edges v10v9, v9v8, v8v7, v7v6.

Let G0 be the graph obtained from G by replacing the path P = (v1, v10, v9, v8, v7, v6) by a

single edge v1v6. Moreover, let `0 be the labelling of G0 such that `0(e) = `(e) for every edge

e ∈ E(G0) ∩ E(G), and `0(v1v6) = `(v1v10). Since ` is equitable and due to the previous remark, `0

is an equitable 2-labelling of G. Now, it suffices to show that `0 is also proper. If this is the case, we arrive at a contradiction since G0 is isomorphic to K3,3 andχΣ(K3,3) = 3 (as proved in [1]).

For the sake of contradiction, suppose that `0 is not proper. Since ` is a proper labelling of G, it follows that c`0(v₁) = c_`0(v₆) in G0 and that these are the only two vertices that are in conflict.

Observe that G0 is a cubic graph and thus, for each v ∈ V (G0), we have c`0(v) ∈ {3, 4, 5, 6}. We

distinguish the following cases: • c`0(v₁) = c_`0(v₆) = 3.

In this case, `0(v4v6) = `0(v2v1) = `0(v3v1) = `0(v5v6) = 1, and so, c`0(v₂), c_`0(v₃), c_`0(v₄),

c`0(v₅) 6= 6. Moreover, so that 3 = c_`0(v₁) /∈ {c_`0(v₂), c_`0(v₃)} and 3 = c_`0(v₆) /∈ {c_`0(v₄),

c`0(v₅)}, we have that c_`0(v₂), c_`0(v₃), c_`0(v₄), c_`0(v₅) ∈ {4, 5}. By symmetry, let us assume

that c`0(v₂) = 5. Then, we must have c_`0(v₄) = c_`0(v₅) = 4 which means `0(v₃v₄) = 1 and

that we must have c`0(v₃) = 5, which is impossible since `0(v₃v₄) = 1.

• c`0(v₁) = c_`0(v₆) = 4.

First, let us assume that `0(v1v6) = 2. In this case, `0(v4v6) = `0(v2v1) = `0(v3v1) = `0(v5v6) =

1, and so, c`0(v₂), c_`0(v₃), c_`0(v₄), c_`0(v₅) 6= 6. Moreover, so that 4 = c_`0(v₁) /∈ {c_`0(v₂), c_`0(v₃)}

and 4 = c`0(v₆) /∈ {c_`0(v₄), c_`0(v₅)}, we have that c_`0(v₂), c_`0(v₃), c_`0(v₄), c_`0(v₅) ∈ {3, 5}. By

symmetry, let us assume that c`0(v₂) = 5. Then, we have c_`0(v₄) ∈ {4, 5}, which conflicts with

either v2 or v6.

Second, let us assume that `0(v1v6) = 1. In this case, we may assume by symmetry that